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The problem requires you to prove: $x^2 + xy + y^2 > 0$ assuming that $x$ and $y$ are not both zero

I made use of a property proven earlier ($x^3 - y^3) = (x-y)(x^2 + xy + y^2)$) and rewrote $x^2 + xy + y^2$ as $\dfrac{x^3 - y^3}{x-y}$. Then I simply tested for the three cases $x = y, x > y, x < y$

For the last two, it's easy to show that the condition holds. But when I consider $x = y$ then the statement that I got evaluates to zero, which is clearly not greater than zero, even though the statement $x^2 + xy + y^2$ should not be zero.

Edit: One stupid mistake. $x = y$ is undefined as it requires division by zero.
What do I do from here?

Bumblebee
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Aayush Agrawal
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6 Answers6

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If $x+y$, then your equation simply says that $0 = 0\cdot (x^2 + xy + y^2)$ which is correct, even if the expression $x^2+xy+y^2$ is not equal to zero.

The equation in this case is correct, but also useless. However, if you know that $x=y$, then the expression simplifies to $$x^2+x\cdot x + x^2$$ which is greater than zero because it is equal to $3x^2$.

5xum
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  • Does that mean that the proof itself is correct? – Aayush Agrawal Jan 13 '16 at 09:25
  • @AayushAgrawal I don't know because for $x>y$ and $x<y$, you haven't written the proof so I cannot tell you if it is correct or not. Also, for $x=y$, you still need to write a sentence or two about how you know that $x^2+xy+y^2>0$. – 5xum Jan 13 '16 at 09:27
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If $x \neq 0 \Rightarrow x^2+xy+y^2 = \left(y+\dfrac{x}{2}\right)^2+\dfrac{3x^2}{4} \geq \dfrac{3x^2}{4}>0$. Similarly you can do the case $y \neq 0$.

DeepSea
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Everyone has given a possible solution but I think they may not have answered your specific question. How to deal with the case $x=y$?.

For $x=y$ you simple look at the expression given $x^2+xy+y^2=3x^2 >0$ when $x \neq 0$. Hence your proof will be complete.

For a simple argument please look at my comment where I have given a link to the same question and the solution I had submitted for it.

Anurag A
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Indeed, for $x\ne y$,

$$x^2+xy+y^2=\frac{x^3-y^3}{x-y}>0$$

as the numerator and denominator have the same sign.

For $x=y$, the remarkable identity cannot be used as it leads to an indeterminate form, and instead

$$x^2+xx+x^2>0,$$

unless $x=y=0$, which is false by hypothesis.

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Seems obvious: we know $(x+y)^2>0$, and that just expands to $x^2+2xy+y^2>0$. So either $xy>0$ to begin with, or else, if $xy<0$, then positive $x^2+y^2$ dominates negative $2xy$. And if it dominates $2xy$, then it certainly dominates $xy$.

John Forkosh
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Hint: $$(x+y)^2-xy=x^2+xy+y^2=(x-y)^2+3xy.$$

Vincenzo Oliva
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